Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-x+y &= -2 \\ -8x+3y &= -6\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-8x = -3y-6$ Divide both sides by $-8$ to isolate $x$ $x = {\dfrac{3}{8}y + \dfrac{3}{4}}$ Substitute this expression for $x$ in the first equation. $-({\dfrac{3}{8}y + \dfrac{3}{4}}) + y = -2$ $-\dfrac{3}{8}y - \dfrac{3}{4} + y = -2$ Simplify by combining terms, then solve for $y$ $\dfrac{5}{8}y - \dfrac{3}{4} = -2$ $\dfrac{5}{8}y = -\dfrac{5}{4}$ $y = -2$ Substitute $-2$ for $y$ in the top equation. $-x- 2 = -2$ $-x-2 = -2$ $-x = 0$ $x = 0$ The solution is $\enspace x = 0, \enspace y = -2$.